What is the foci (x^2)/4-((y+1)/(16))^2=1 ?

The foci (x^2)/4-((y+1)/(16))^2=1 is (2sqrt(65),-1),(-2sqrt(65),-1)

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foci (x^2)/4-((y+1)/(16))^2=1

foci

\frac{x ^{2}}{4} - (\frac{y + 1}{16})^{2} = 1

(265, - 1), (- 265, - 1)

Solution steps

(h + c, k), (h - c, k)

Calculate Hyperbola properties

\frac{x ^{2}}{4} - (\frac{y + 1}{16})^{2} = 1 :

Right-left Hyperbola with

(h, k) = (0, - 1), a = 2, b = 16

(0 + c, - 1), (0 - c, - 1)

Compute

c :

c = 2^{2} + 1 6^{2} : 265

(0 + 265, - 1), (0 - 265, - 1)

Refine

(265, - 1), (- 265, - 1)

\frac{x ^{2}}{36} - \frac{y ^{2}}{2} = 1

x^{2} - 4 y^{2} + 16 y - 32 = 0

3 = x^{2} - y^{2}

4 x^{2} - 9 y^{2} - 8 x + 72 y - 176 = 0

- x^{2} + 4 y^{2} - 8 y - 12 = 0

What is the foci (x^2)/4-((y+1)/(16))^2=1 ?
The foci (x^2)/4-((y+1)/(16))^2=1 is (2sqrt(65),-1),(-2sqrt(65),-1)